NotThatSmart

Haha, this is a very funny problem and shows the world of fractions. 

a) When x = 10, we have \(\frac{2(10)+3}{2(10)+3} = \frac{23}{23} = 1\). So yes, when x = 10, we have the first expression is equal to the second expression. 

b) Ok, so then why doesn't \(-\frac{3}{2}\) work? It looks like the numerator and denominator are equal, so it's always 1 right? 

Well, let's simplify and go from there. We have \(\frac{2(-2/3)+3}{2(-2/3)+3}=\frac{-3+3}{-3 + 3}=0/0\). WAIT a minute, the denominator can't be 0! That would make the value undefined, not 1!

c) Ok, so the reaon why the answer is not 1 in part (b) is because the denominator would become 0. 

However, as long as it doesn't equal 0, the numerator and denominator will always be equal, meaning the value of the fraction will always be 1!

I hope I answered your question!

Thanks!

Hey Akhain, 

Can you please explain to me how you found the contraint 30x2​+305y2​=1? 

My puny brain is not exactly sure I understand that part....

Thanks!

Let  \(x\mathbin{\spadesuit}y = x^2/y\) for all \(x\) and \(y\) such that \(y\neq 0\). Find all values of \(a\) such that \(a \mathbin{\spadesuit} (a + 1) = 9\). List the values you find in increasing order, separated by commas.

Alright, so in this case, we have \(x = a\) and \(y = a +1 \). Putting into the form of \(x\mathbin{\spadesuit}y = x^2/y\), we have \(a \mathbin{\spadesuit} (a + 1) = \frac{a^2}{a+1} = 9\). 

Multiplying both sides by a + 1 and distributing the 9 in, we have \(a^2 = 9a + 9\). Now we can write the quadratic \(a^2 - 9a - 9 = 0\). 

Unforunately, we can't factor this directly, so we have to use the quadratic formula to find that

\(a=\frac{3\sqrt{13}+9}{2}\\ a=\frac{-3\sqrt{13}+9}{2}\)

This means our final answer is \(a=\frac{3\sqrt{13}+9}{2}, \frac{-3\sqrt{13}+9}{2}\)

Thanks!

This is a pretty easy problem to solve, set I'll try my best to explain it!

We know the area of a circle is \(r^2\pi \). This is when we have the full 360 degrees. However, for this problem, we have the fact that only 83 degrees is shown. 

This is easily solvable, as that means that 83/360 of the total area has been calculated. Please understand why this is. 

So, we can find the radius with the equation \(\frac{83}{360}r^2\pi = 60\pi\). Canceling out the pi and isolating r^2, we have \(r^2= 60 \cdot \frac{360}{83}\). 

This is where it gets a bit messy. I won't show all my work, but just know that after canceling everything out, we have \(r^2 = \frac{21600}{83}\). Squaring rooting both sides we have \(r=\frac{60\sqrt{498}}{83}\). 

Thanks!

This is actually quite an easy question!

At first, it might look like we have to distribute out and actually go through and raise 3x^2 + 11 to the power of 12, which is insanely time consuming and honestly, just not worth it. 

However, there is a simple trick we could use to solve this problem. Notice that the problem is asking for the degree of polynomial, which is the highest power in this polynomial. 

This means we wouldn't have to actually find the entire polynomial. 

Now, let's note something. We see the orinal polynomial has degree 2, since it's leading term has a power of 2. This means that not matter what, it is not possible to get a higher degree than whatever this orginal degree is raised to the power of. What I mean is, when we raise 3x^2 to the power of 12, the other terms in the polynomial will not have a higher power than that! 

So the highest degree is just (x^2)^12, which is just x^24. So the new polynomial has degree 24!

I hope you understood my explenation!

Thanks!

This problem looks really complicated and a nightmare to solve, but it actually is quite simple!

First, we have to factor in the 4m-2 into the parenthesis, getting us \(\frac{{1}^{4m-2}}{{49}^{4m-2}} = {7}^{-10m-2}\). 

This still looks far off from our goal of finding m, but notice that 49 is just 7^2 and the numerator is just 1. This means that we just have \(\frac{1}{{7}^{2(4m-2)}}\). 

Alright, now we're closer to comparable levels. However, we have once term in the denominator and one in the numerator. What should we do about that?

Well, the answer lies in the fact that \({a}^{-1} = \frac{1}{a}\). If we factor out -1 from the power in the right side of the equation, we have \({7}^{-(10m+2)}\), meaning that we technically just have the final equation \(\frac{1}{{7}^{8m-4}} = \frac{1}{{7}^{10m+2}}\). In order for the equation to be satisfied, we must have \({7}^{8m-4} = {7}^{10m+2}\), meaning that \(8m - 4 =10m + 2\). 

Combining like terms gets us \(-2m = 6\) so \(m = -3\)!

The key to this problem is to realize that we have to kind of force our way to a similar expression on both sides so that we could write an equation. I hope I answered your question correctly and clearly!

Thanks!